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JEE Advanced 2015Paper 1
Exam
JEE Advanced
Year
2015
Paper
Paper 1
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Advanced 2015 · Paper 1 · Mathematics · Three Dimensional Geometry

In R ^ 3 , let L be a straight line passing through the origin....

In R3{R}^{3} , let L be a straight line passing through the origin. Suppose that all the points on L are at a constant distance from the two planes P1:x+2y−z+1=0{P}_{1}: x + 2y - z + 1 = 0 and P2:2x−y+z−1=0.{P}_{2}: 2x - y + z - 1 = 0 . Let M be the locus of the feet of the perpendiculars drawn from the points on L to the plane P1{P}_{1} . Which of the following points lie (s) on M ?

Options

  1. A(0,−56,−23)\left(0, - \dfrac{5}{6}, - \dfrac{2}{3}\right)
  2. B(−16,−13,16)\left(- \dfrac{1}{6}, - \dfrac{1}{3}, \dfrac{1}{6}\right)
  3. C(−56,0,16)\left(- \dfrac{5}{6}, 0, \dfrac{1}{6}\right)
  4. D(−13,0,23)\left(- \dfrac{1}{3}, 0, \dfrac{2}{3}\right)

Correct answer

A. (0,−56,−23)\left(0, - \dfrac{5}{6}, - \dfrac{2}{3}\right)

Step-by-step solution

Line L will lie on one of the angle bisector planes and will be parallel to line of intersection of the given planes. Also locus of foot of perpendicular of line L in plane P1{P}_{1} will be line M, which will be parallel L.
Foot of perpendicular from (0, 0, 0) on plane P1{P}_{1} is \left(- \dfrac{1}{6}, - \dfrac{1}{3}, \dfrac{1}{6}\right)<br>HenceequationoflineM<br><br >Hence equation of line M<br >\dfrac{x + \dfrac{1}{6}}{1} = \dfrac{y + \dfrac{1}{3}}{- 3} = \dfrac{z - \dfrac{1}{6}}{- 5}<br>Where<br >Where\overset{^}{i} - 3\overset{^}{j} - 5\overset{^}{k}isvectorparalleltolineofintersectionofis vector parallel to line of intersection of{P}_{1}andand{P}_{2}.<br>Onchecking(0,−56−23),.<br >On checking \left(0, - \dfrac{5}{6} - \dfrac{2}{3}\right),\left(- \dfrac{1}{6}, - \dfrac{1}{3}, \dfrac{1}{6}\right) satisfy the above equation.
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